An application of the pigeonhole principle on a $7\times 7$ board Before moving on to my main question, I would like to write the basic question that I know the solution method well.
Basic Question: Prove that no matter how $50$ points are chosen from inside or over a square with a side length of $7$ units, there are two points with a distance is $\leq \sqrt{2}$ between them.
Solution: Let's split the board in $7\times 7$ type. So we have $49$ unit squares. According to the pigeonhole principle, $ \left\lfloor\dfrac{50}{49}\right\rfloor + 1 = 2$ points are in or on the same square. The distance between them is $\leq \sqrt{2}$.
At this stage, the following question came to my mind:
Main Question: No matter how $n$ points are taken from inside (or over) a square with side lengths of $7$ units, there can always be a pair of points whose distance is $\leq \sqrt{2}$ unit. What is the smallest value of $n$ that makes this condition possible?
At first I thought the answer would simply be $50$. I tried to place the points as close to each other as the distance between them was slightly greater than $\sqrt{2}$. I got $32$ points. So I feel that $n = 32+1=33$ is the minimum desired value. I have no rigorous proof.
Thank you for your advice and assistance.
 A: The problem of arranging points in a square with no two points near each other is equivalent to the well-studied problem of packing circles in a square.  Specifically, if you choose $n$ points in a $7 \times 7$ square with no two points within $\sqrt 2$ units of each other, then the circles of radius $\frac{\sqrt 2}{2}$ centered on the given points do not intersect and are all contained in a square of side length $7 + \sqrt 2$.  The converse holds as well.
As a result, after rescaling distances, you are asking how many unit circles can be packed in a square of side length $(7 + \sqrt 2) \cdot \sqrt 2 = 7 \sqrt 2 + 2 \approx 11.8995$.  According to here, the answer to this is 35:  the tightest possible packing of $36$ unit circles is in a square of side length $12$ (in a square grid), while $35$ unit circles can be packed in a square of side length at most $s \approx 11.8637$ (picture here, not known to be optimal).  This corresponds to a packing of $35$ points in a square of side length $\frac{s-2}{\sqrt 2} \approx 6.9747$ with all distances at least $\sqrt 2$ (or a slightly larger square with all distances greater than $\sqrt 2$).  Meanwhile, for $36$ points you would need a square of side length at least $\frac{12 - 2}{\sqrt 2} \approx 7.071$.
